Birthday Probability

Number of people N:           Probability some duplication= =(probability not all different dates)

    

Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Probability that all k random selections with replacement from n are different i.e. no duplicates is:
Probability that they are NOT all different is 1- this value.

Number of letters N:           Probability some duplication= =(probability not all different letters)

Number of strings of N letters:     

Number of strings with no dups*:


Probability that a sample selected with replacement of size n from a population of size N has no duplicates is N!/(N-n)!Nn = N·(N-1)·...·(N-n+1) / Nn = Product(k=1 to n-1)(1-k/N)

N n: 10 30 100
1000 P(no dups) 0.955861 0.644461 0.005959
10000 P(no dups) 0.995509 0.957392 0.608566
100,000 P(no dups) 0.999550 0.995659 0.951690
1,000,000 P(no dups) 0.999955 0.999565 0.995062

Probability the sample has duplicates is 1-P(no dups).